Applications of Liouville's theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T07:31:07Z http://mathoverflow.net/feeds/question/100750 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100750/applications-of-liouvilles-theorem Applications of Liouville's theorem js 2012-06-27T08:00:16Z 2012-06-27T18:42:45Z <p>I'm looking for "nice" applications of Liouville's theorem (every bounded entire map is constant) <em>outside</em> the area of complex analysis.</p> <p>An example of what I'm <em>not</em> looking for : a non-constant entire function has dense image (this is essentially a corollary).</p> <p>An example of the kind of thing I'm looking for : a complex matrix whose conjugacy class is bounded must be a homothety (if $A$ is such a matrix and $B$ is an other matrix, then $z \mapsto e^{-z B} A e^{z B}$ is entire and bounded hence constant, but its derivative at $0$ is $[A,B]$ : thus $[A,B]=0$). In a similar vein : a subalgebra of $M_n (\mathbb{C})$ on which the spectral radius is submultiplicative is simultaneously triangularizable.</p> http://mathoverflow.net/questions/100750/applications-of-liouvilles-theorem/100753#100753 Answer by Marc Palm for Applications of Liouville's theorem Marc Palm 2012-06-27T08:18:05Z 2012-06-27T08:18:05Z <p>The fundamental theorem of algebra: The field $\mathbb{C}$ is algebraically closed.</p> <p>See here: <a href="http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra#Complex-analytic_proofs" rel="nofollow">http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra#Complex-analytic_proofs</a></p> http://mathoverflow.net/questions/100750/applications-of-liouvilles-theorem/100758#100758 Answer by Yemon Choi for Applications of Liouville's theorem Yemon Choi 2012-06-27T09:08:23Z 2012-06-27T09:08:23Z <p>Fuglede's theorem (if $N$ and $P$ are commuting operators on Hilbert space, and $N$ is normal, then $P$ commutes with $N^*$) has a slick proof using the vector-valued Liouville theorem. I guess the special case of matrices may be sufficiently interesting to be included as an exercise. (For normal matrices one could just use the spectral theorem, I guess.)</p> http://mathoverflow.net/questions/100750/applications-of-liouvilles-theorem/100760#100760 Answer by Salvo Tringali for Applications of Liouville's theorem Salvo Tringali 2012-06-27T09:35:18Z 2012-06-27T09:42:58Z <p>The first published proof of the Mazur-Gelfand theorem, due to Gelfand himself (though previously announced without proof by Mazur), is based on the vector-valued version of the Liouville theorem, which was further extended by Arens to cover a more general situation (see [1] and references therein).</p> <p>[1] R. Arens (1947), <em>Linear topological division algebras,</em> Bull. AMS, Vol. 53, pp. 623-630.</p> http://mathoverflow.net/questions/100750/applications-of-liouvilles-theorem/100762#100762 Answer by diverietti for Applications of Liouville's theorem diverietti 2012-06-27T10:24:59Z 2012-06-27T12:21:01Z <p>A very nice application of Liouville's theorem in functional analysis is the following, which is of great theoretical and practical importance. </p> <p><strong>Theorem (Spectrum).</strong> <em>If $X\ne\lbrace0\rbrace$ is a complex Banch space and $T\colon X\to X$ a bounded linear operator, then its spectrum $\sigma(T)\ne\emptyset$.</em> </p> <p>First of all, let $X$ a complex Banach space, $B(X,X)$ the space of bounded linear operators from $X$ to $X$ and $\Lambda\subset\mathbb C$ a domain of the complex plane. Consider a function $$S\colon\Lambda\to B(X,X), \qquad\lambda\mapsto S_\lambda.$$</p> <p><strong>Definition.</strong> The map $S$ is said to be <em>holomorphic</em> on $\Lambda$ if for every $x\in X$ and $f\in X^*$ the function $h$ defined by $$h(\lambda)=f(S_\lambda(x))$$ is holomorphic at every $\lambda_0\in\Lambda$.</p> <p>The following proposition is an easy exercise.</p> <p><strong>Proposition (Holomorphy of $R_\lambda$).</strong> <em>The resolvent</em> $R_\lambda(T)$ <em>of a bounded linear operator</em> $T\in B(X,X)$ <em>is holomorphic at every point of the resolvent set</em> $\rho(T)$ <em>of</em> $T$.</p> <p>The proof of the Spectrum theorem is then quite elementary and goes as follows. </p> <p><em>Proof.</em> By assumption, $X\ne\lbrace0\rbrace$. If $T=0$, then $\sigma(T)=\lbrace0\rbrace\ne\emptyset$. So, let $T\ne 0$ and $$R_\lambda=(T-\lambda I)^{-1}=-\frac 1\lambda\sum_{j=0}^\infty(\frac 1\lambda T)^j.$$ This series is convergent for all $|\lambda|>||T||$, and thus it converges absolutely for instance for $|\lambda|>2||T||$. For these $\lambda$, by the formula for the sum of a geometric series, we have $$||R_\lambda||\le\frac 1{||T||}.$$ If $\sigma(T)=\emptyset$, then by definition the resolvent $\rho(T)$ is the whole complex plane. Hence, $R_\lambda$ is holomorphic for all $\lambda$. Consequently, for a fixed $x\in X$ and $f\in X^*$, the function $h$ defined by $$h(\lambda)=f(R_\lambda(x))$$ is holomorphic on $\mathbb C$, that is, it is an entire function. Now, $h$ is in particular continuous and thus bounded on the compact disk $|\lambda|\le 2||T||$. But $h$ is also bounded for $\lambda\ge 2||T||$ since $||R_\lambda||\le1/||T||$ and $$|h(\lambda)|=|f(R_\lambda(x))|\le||f||\cdot||R_\lambda(x)||\le||f||\cdot||R_\lambda||\cdot||x||\le\frac{||f||\cdot||x||}{||T||}.$$ Hence, $h$ is constant by <em>Liouville's theorem</em>. But this implies that $R_\lambda$ is independent of $\lambda$ and that so is $R_\lambda^{-1}=T-\lambda I$, which is a contradiction.$\qquad\square$ </p> <p>Observe that in the finite dimensional case, that is $X=\mathbb C^n$, the Spectrum Theorem says that the characteristic polynomial $\det(A-\lambda I)$ of a complex $(n\times n)$-matrix $A$ has a solution, which is just the fundamental theorem of algebra, which in turn follows again by Liouville's theorem... </p> http://mathoverflow.net/questions/100750/applications-of-liouvilles-theorem/100778#100778 Answer by Steven Gubkin for Applications of Liouville's theorem Steven Gubkin 2012-06-27T13:54:50Z 2012-06-27T18:42:45Z <p>There are many cool applications when combined with the uniformization theorem. Not sure if you count them as "complex analysis" or not - you could really think of them as algebraic geometry. For example:</p> <p>The only meromorphic functions $f$ and $g$ satisfying $f^n+g^n = 1$ for $n>3$ are constant. </p> <p>Proof sketch: If there were such functions they would define a map $F$ from the plane to the Riemann surface $x^n+y^n = 1$. We can compute the genus of this guy using Hurwitz - it is $(n-1)(n-2)/2$. So for $n>3$ the genus is bigger than two - the corresponding Riemann surface has negative curvature, and by uniformization it has the disk as its holomorphic universal cover. But then, $F$ would factor through the disk, and so by Louville F had to be constant.</p> <p>Note:<br> for n=1 the equation has lots of solutions</p> <p>for n=2 sine and cosine work (for instance)</p> <p>for n=3 the Riemann surface has genus 1, and so its universal cover is the plane. You can find explicit solutions to the equation by using theta functions.</p> <p>EDIT: another very important example of the same sort of reasoning is the little Picard theorem. At a high level, the proof just says that the plane with 2 points deleted has a holomorphic universal covering by the disk, so any entire function which misses 2 points factors through the disk - Louville gives the contradiction.</p>